A body is released from the top of a tower H | vedclass

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(C) Let the total height of the tower be $H$. The time taken to reach the ground is $t$. Using the equation of motion $s = ut + \frac{1}{2}at^2$,where

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$3\frac{H}{4}$ metre from ground

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(C) Let the total height of the tower be $H$. The time taken to reach the ground is $t$. Using the equation of motion $s = ut + \frac{1}{2}at^2$,where $u = 0$ and $a = g$:$H = \frac{1}{2}gt^2$  --- $(i)$After time $\frac{t}{2}$,the distance $x$ covered by the body from the top is:$x = \frac{1}{2}g(\frac{t}{2})^2 = \frac{1}{2}g(\frac{t^2}{4}) = \frac{1}{8}gt^2$ --- $(ii)$From $(i)$,we know $\frac{1}{2}gt^2 = H$,so $\frac{1}{4}(\frac{1}{2}gt^2) = \frac{H}{4}$.Thus,$x = \frac{H}{4}$ from the top.The height of the body from the ground is $H - x = H - \frac{H}{4} = \frac{3H}{4}$ metres.

$A$ body is released from the top of a tower $H$ metre high. It takes $t$ seconds to reach the ground. The height of the body $\frac{t}{2}$ seconds after release is

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